Aradioactive nucleus alpha decays to yield a sodium-24 nucleus in 14.8 hours. what was the identity of the original nucleus?

2813Al

Explanation:

You know that alpha decay takes place when a α-particle is being ejected from the nucleus of a radioactive isotope.An 

α-particle is simply the nucleus of a helium-4 atom. A helium-4 atom has a total of two protons and two neutrons in its nucleus, and two electrons surrounding that nucleus.

In that case, if a α-particle is the nucleus of a helium-4 atom, then it must have a mass number equal to 4, since it has two protons and two neutrons, and a net charge of (2+) since it no longer has the two electrons that the helium-4 atom has.

So, you know that an unknown radioactive isotope decays via alpha decay to yield a sodium-24 nucleus.

A sodium-24 nucleus contains 11 protons and 13 neutrons. This means that you can write

AZX→2411Na+42α

Here A and Z represent the unknown element's mass number and atomic number, respectively.

So, if you take into account the fact that mass number and atomic number must be conserved in a nuclear reaction, you can say that

A=24+4  and  Z=11+2

The unknown isotope will thus have

{A=28Z=13

A quick look at the periodic table will show you that the element that has 13 protons in its nucleus is aluminum, Al. This means that you're dealing with aluminum-28, an isotope of aluminum that has 15 neutrons in its nucleus.

The complete nuclear equation will be

2813Al→2411Na+42α

The answer is c. 320 g

The density of water is 1 g/cm cubed and in order for the pebble to sink its density has to be greater than the density of water and in order for the twig to float its density has to be less than the density of water